Problem: What is the value of the following logarithm? $\log_{8} 2$
If $b^y = x$ , then $\log_{b} x = y$ Notice that $2$ is the cube root of $8$ That is, $\sqrt[3]{8} = 8^{1/3} = 2$ Thus, $\log_{8} 2 = \dfrac{1}{3}$.